Physics And Measurement Question 49

Question: The dimensions of universal gravitational constant are

[ CBSE PMT 1988, 92; 2004 ]

Options:

A) $ {{M}^{-2}}L^{2}{{T}^{-2}} $

B) $ {{M}^{-1}}L^{3}{{T}^{-2}} $

C) $ M{{L}^{-1}}{{T}^{-2}} $

D) $ ML^{2}{{T}^{-2}} $

Show Answer

Answer:

Correct Answer: B

Solution:

$ F=\frac{Gm _{1}m _{2}}{d^{2}}\Rightarrow G=\frac{Fd^{2}}{m _{1}m _{2}} $

$ 1ly=9.46\times 10^{15}meter $

$ [G]=\frac{[ML{{T}^{-2}}][L^{2}]}{[M^{2}]}=[{{M}^{-1}}L^{3}{{T}^{-2}}] $



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